October 2024 - Solution
October 2024 Solution:
The numerical solutions are
X = 1817198712146313216, A = 1817198712, B = 146313216 X = 8419411123272236924928, A = 84194111232, B = 72236924928
The solution is based on a search, which can be optimized in various ways. We give the method described by Alex Izvalov (thanks, Alex!):
- As the number X should contain all the digits 1,2,3,4,6,7,8,9 at least once, and B is the product of digits of X, then B = 2^{a+7}\cdot3^{b+4}\cdot7^{c+1} If B is a perfect square, then it should be in the form: B = 2^{2a+8}\cdot3^{2b+4}\cdot7^{2c+2}, where a,b,c are natural numbers
To search for B in the (almost) increasing order we'll loop through the sum of a,b,c, s = a+b+c and then we'll loop a from 0 to s, b from 0 to s-a, and then c = s-a-b
Next we should check if B contains digits 0 or 5, and we'll proceed to the next cycle iteration in this case
As X is a multiple of B, then (let m be the length of B) A\cdot 10^{m}+B is a multiple of B, then prime factoring of A should have as many 3s and 7s as B has, and the number of 2s should be not less than the number of 2s in B minus the length of B. We'll call the minimal number consisting of these prime factors, A_0
As now we know the product of digits of X, and we know B, we'll know the product of digits of A, too (and for the advanced problem we can break the loop early if it's not a perfect cube).
We'll use several tactics to reduce the field of search.
Then we can proceed in 2 directions: either search for A as a multiple of A_0\cdot q and loop through q and then check the digits, or construct A as a number of gradually increasing length which product of numbers should be equal to \frac{B}{\text{prod}(B)}, and then check if it's a multiple of A_0.
I found the first, more straightforward way, easier to use when searching for the candidates for the minimal X, and the second way to prove that the results are indeed minimal.